Math Problem Statement
a. Use the Intermediate Value Theorem to show that the following equation has a solution on the given interval.
b. Use a graphing utility to find all the solutions to the equation on the given interval.
c. Illustrate your answers with an appropriate graph.
3 x cubed plus x minus 1 equals 0; (0,1)
Question content area bottom
Part 1
a. The Intermediate Value Theorem states that if f is
continuous
on the interval
[a,b]
and L is a number
strictly between
f(a) and f(b), then
there exists at least one
number c in
(a,b)
satisfying
f left parenthesis c right parenthesis equals L.
Part 2
For which values of x is the function f(x)equals3 x cubed plus x minus 1 continuous?
A.
It is continuous for some x, but not on [0,1].
B.
It is continuous for all x.
Your answer is correct.C.
It is continuous on [0,1], but not for all x.
D.
It is not continuous on any interval.
Part 3
Evaluate the function f(x) at the left endpoint.
The value of the function at the left endpoint is
negative 1.
(Type an integer or decimal rounded to three decimal places as needed.)
Part 4
Evaluate the function f(x) at the right endpoint.
The value of the function at the right endpoint is
3.
(Type an integer or decimal rounded to three decimal places as needed.)
Part 5
Why can the Intermediate Value Theorem be used to show that the equation has a solution on (0,1)?
A.
It can be used because f(x)equals3 x cubed plus x minus 1 is continuous on [0,1] and the function is defined at xequals0 and xequals1.
B.
It can be used because f(x)equals3 x cubed plus x minus 1 is continuous on [0,1] and 0 is between f(0) and f(1).
Your answer is correct.C.
It can be used because f(x)equals3 x cubed plus x minus 1 is defined on (0,1) and 0less thanf(0)less thanf(1).
D.
It can be used because f(x)equals3 x cubed plus x minus 1 is defined on (0,1) and f(0)less thanf(1)less than0.
Part 6
b. There is/are a solution(s) to the equation in (0,1) at xalmost equals
enter your response here.
(Type an integer or decimal rounded to three decimal places as needed. Use a comma to separate answers as needed.)
Solution
Let's break down the problem step by step and address each part.
Part 1: Using the Intermediate Value Theorem (IVT)
The Intermediate Value Theorem states that if a function is continuous on the interval , and is a number between and , then there exists at least one number in such that .
We are given the equation: and the interval .
Part 2: Continuity of
The function is , which is a polynomial function. Polynomial functions are continuous for all real numbers, meaning the correct answer is:
Answer: B. It is continuous for all .
Part 3: Evaluate at the left endpoint (0)
We need to evaluate the function at : So the value of the function at the left endpoint is .
Part 4: Evaluate at the right endpoint (1)
Next, we evaluate the function at : So the value of the function at the right endpoint is .
Part 5: Why can the Intermediate Value Theorem be used?
We can use the Intermediate Value Theorem because:
- The function is continuous on the interval .
- The function takes values and .
- The number is strictly between and .
Thus, the correct answer is: Answer: B. It can be used because is continuous on and 0 is between and .
Part 6: Find the solution in the interval (0, 1)
We are looking for a value in such that . From the IVT, since and , and since lies between and , there must be at least one in the interval such that .
We can use a graphing utility to find the exact solution(s) to the equation in the interval .
After graphing the function or solving numerically, the solutions to the equation in the interval are approximately:
Summary of Answers:
- The function is continuous for all .
- The value of the function at is .
- The value of the function at is .
- The IVT can be used because lies between and .
- The solution to the equation in the interval is approximately .
Would you like a graph of the function to visually verify the solution?
If you have any questions or need more details, feel free to ask!
Additional Questions:
- How do you find other solutions to the equation outside the interval ?
- What are other methods (besides graphing) to solve polynomial equations?
- How would you apply the IVT to a non-polynomial function, like a trigonometric one?
- Can a function be continuous on an interval but still not have a solution for ?
- How do you estimate solutions to equations numerically if the exact value isn't easily obtainable?
Tip: When applying the IVT, always ensure the function is continuous over the entire interval and that the number you're checking lies between the values of the function at the endpoints.
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Math Problem Analysis
Mathematical Concepts
Intermediate Value Theorem
Polynomial Functions
Continuity
Graphing Functions
Formulas
f(x) = 3x^3 + x - 1
Theorems
Intermediate Value Theorem
Suitable Grade Level
Grades 10-12